An exothermic chemical reaction is being performed in a CSTR, and the temperature is beingcontrolled by adjusting the water flow rate through a cooling jacket. At the design conditions,the cooling water flow rate is 50 gpm and the cooling water pressure drop across the jacket is20 psi. The cooling water pressure drop across the jacket is proportional to the flow ratesquared.An equal percentage control valve is used with R = 20. The combined pressure drop across thecontrol valve and cooling jacket is constant at 50 psi.(a) Should the control valve be fail-open or fail-closed?(b) Determine the C- for the valve if the valve is 40% open (e.g., f(€) = 0.40) at the designconditions.(c) Calculate the maximum cooling water flow rate through the control valve.

To control fluid flow we use control valves. There will be control signal to open or close the valve. These are placed in plant for fail safe operation. These control over pressure ,temperature and liquid level conditions.The control valve at exothermic feed location must be air open valve or fail to close valve .These valves open when given a control signal. This air open valve increases the opening with increase in pressure. The control valve at coolant location must be air close valve or fail to open valve . In this air pressure increase the valve gets more closed. The design flow rate is 50gpm We can derive that ∆Pc=20Psi∆Pc=20F502∆Pc=(8×10-3)F2 Calculation of control valve f(l)=0.4∆Pv+∆Pc=50Psi∆Pv=50Psi-∆Pc∆Pv=50Psi-20Psi∆Pv=30PsiF=Cvf(l)∆PvSpecifc gravity of waterCv=Ff(l)∆PvSpecifc gravity of waterCv=500.4301Cv=22.82 Maximum cooling water flow rate ∆Pc=(8×10-3)F2∆PT=50Psi∆PT=∆Pv+∆Pc∆Pv=∆PT-∆Pc∆Pv=50-(8×10-3)F2F=22.82(0.4)(50-(8×10-3)F2)F=9.128(50-(8×10-3)F2) F9.128=50-(8×10-3)F2F9.1282=50-(8×10-3)F2F9.1282+(8×10-3)F2=50F29.1282+(8×10-3)F2=50F219.1282+8×10-3=50F2(0.0200)=50F2=2499.76F=2499.76F=49.99 gpm Final answers a) Coolant control valve must be air close or fail to open valve b)Cv=22.82 c)49.99 gpm is the maximum coolant flow rate